Respuesta :
Answer:
1. [tex]P(X>69)=P(Z>-1.136)=1-P(Z<-1.136)=0.872[/tex]
2. [tex]P(X<81)=P(Z<0.227)=0.590[/tex]
3. [tex]P(52.6 \leq X \leq 105.4)=P(-3<Z<3)=P(Z<3)-P(Z<-3)=0.999-0.00135=0.9973[/tex]
Step-by-step explanation:
Assuming the value for the mean 79 and the deviation 8.8
1) Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
2) Part a
Let X the random variable that represent the amount of collagen of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(79,8.8)[/tex]
Where [tex]\mu=79[/tex] and [tex]\sigma=8.8[/tex]
We are interested on this probability
[tex]P(X>69)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X>69)=P(\frac{X-\mu}{\sigma}>\frac{69-\mu}{\sigma})=P(Z<\frac{69-79}{8.8})=P(Z>-1.136)[/tex]
And we can find this probability on this way:
[tex]P(Z>-1.136)=1-P(Z<-1.136)=0.872[/tex]
3) Part b
[tex]P(X<81)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X<81)=P(\frac{X-\mu}{\sigma}<\frac{81-\mu}{\sigma})=P(Z<\frac{81-79}{8.8})=P(Z<0.227)[/tex]
And we can find this probability on this way:
[tex]P(Z<0.227)=0.590[/tex]
4) Part c
We need to find first the limits for 3 deviations within the mean
[tex]\mu -3 \sigma = 79-3(8.8)=52.6[/tex]
[tex]\mu +3 \sigma = 79+3(8.8)=105.4[/tex]
[tex]P(52.6<X<105.4)=P(\frac{52.6-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{105.4-\mu}{\sigma})=P(\frac{52.6-79}{8.8}<Z<\frac{105.4-79}{8.8})=P(-3<Z<3)[/tex]
And we can find this probability on this way:
[tex]P(-3<Z<3)=P(Z<3)-P(Z<-3)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-3<Z<3)=P(Z<3)-P(Z<-3)=0.999-0.00135=0.9973[/tex]
